Question

Angle between the tangents drawn from $(1,4)$ to the parabola $y^2=4x$ is $\frac{\pi }{m}$, where $m$ is equal to

Answer 1

Equations of the pair of tangents is
$(y^2-4x)(16-4)={[4y-2(x+1)]}^2$
$\Rightarrow$ $12(y^2-4x)=4((2y-x-1){)}^2$
$\Rightarrow$ $3(y^2-4x)=4y^2+x^2+1-4xy+2x-4y$
$\Rightarrow$ $x^2+y^2-4xy+14x-4y+1=0$
If $\theta$ is the angle between these lines
Then $\tan \theta =\frac{2\sqrt{{\left(2\right)}^2-\left(1\right)\left(1\right)}}{\left(1\right)\left(1\right)}=\sqrt{3}$
$\Rightarrow$ $\theta =\frac{\pi }{3}=\frac{\pi }{m}\Rightarrow m=3$.